# What is the slope of the polar curve f(theta) = thetasintheta - cos^3theta + tantheta at theta = pi/3?

Oct 17, 2016

The slope of the tangent is $m \approx 7.7$

#### Explanation:

Given:

$r \left(\theta\right) = \left(\theta\right) \sin \left(\theta\right) - {\cos}^{3} \left(\theta\right) + \tan \left(\theta\right) , \theta = \frac{\pi}{3}$

Tangents with polar coordinates gives us the equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr} \left(\theta\right)}{d \theta} \sin \left(\theta\right) + r \left(\theta\right) \cos \left(\theta\right)}{\frac{\mathrm{dr} \left(\theta\right)}{d \theta} \cos \left(\theta\right) - r \left(\theta\right) \sin \left(\theta\right)}$

(dr(theta))/(d theta) = sin(θ)+θ cos(θ)+sec^2(θ)+3 sin(θ) cos^2(θ)

The slope, m, is:

$m = \frac{\frac{\mathrm{dr} \left(\frac{\pi}{3}\right)}{d \theta} \sin \left(\frac{\pi}{3}\right) + r \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}\right)}{\frac{\mathrm{dr} \left(\frac{\pi}{3}\right)}{d \theta} \cos \left(\frac{\pi}{3}\right) - r \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{3}\right)}$

(dr(pi/3))/(d theta) = 4+(7 sqrt(3))/8+π/6

r(pi/3) = -1/8+sqrt(3)+πsqrt(3)/(6)

$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

$\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$

Substituting the above into the equation for m:

m = ((4+(7 sqrt(3))/8+π/6)sqrt(3)/2 + (-1/8+sqrt(3)+πsqrt(3)/(6))(1/2))/((4+(7 sqrt(3))/8+π/6)(1/2) - (-1/8+sqrt(3)+πsqrt(3)/(6))sqrt(3)/2)

$m \approx 7.7$